Question: Solve for $r$, $ -\dfrac{3}{r - 4} = \dfrac{7}{r - 4} + \dfrac{2r + 9}{5r - 20} $
Explanation: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $r - 4$ $r - 4$ and $5r - 20$ The common denominator is $5r - 20$ To get $5r - 20$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ -\dfrac{3}{r - 4} \times \dfrac{5}{5} = -\dfrac{15}{5r - 20} $ To get $5r - 20$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ \dfrac{7}{r - 4} \times \dfrac{5}{5} = \dfrac{35}{5r - 20} $ The denominator of the third term is already $5r - 20$ , so we don't need to change it. This give us: $ -\dfrac{15}{5r - 20} = \dfrac{35}{5r - 20} + \dfrac{2r + 9}{5r - 20} $ If we multiply both sides of the equation by $5r - 20$ , we get: $ -15 = 35 + 2r + 9$ $ -15 = 2r + 44$ $ -59 = 2r $ $ r = -\dfrac{59}{2}$